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My hypothesis states that "Multiplayer Game is more enjoyable than the single-player game" and lets suppose I have 2 variables one categorical with two categories and the other one is ordinal (Likert scale 1 to 5). Which test should I use to prove my hypothesis?

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  • $\begingroup$ (a) It seems you give no data for the variable with two categories. So, we can comment specifically only on the Likert variable. (b) You can 'test' a hypothesis $H_0$ against an alternative $H_a,$ which may give an idea whether to believe the hypothesis or the alternative, but "prove" is not really the right word to use. $\endgroup$
    – BruceET
    Commented Feb 20, 2022 at 2:30

1 Answer 1

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If you have Likert data as below, for Multi-player and Single-player games, you might try a Wilcoxon rank sum test in R.

multi = c(5,4,5,4,4,4,4)
singl = c(3,2,2,1,1,1,1)
wilcox.test(multi, singl)

    Wilcoxon rank sum test 
    with continuity correction

data:  multi and singl
W = 49, p-value = 0.00147
alternative hypothesis: 
 true location shift is not equal to 0

Warning message:
In wilcox.test.default(multi, singl) :
 cannot compute exact p-value with ties

The warning message is because, for small sample sizes, this test cannot give exact P-values for data with multiple tied observations (as in your data).

However, in your example all of the Multi-player Likert scores are 4 and above and all of the Single-player scores are 3 and below. The probability that out of 14 scores 7 with high values would be in one group and 7 with low values would be in the other group is $2/{14 \choose 7} = 0.0005827506,$ which is the P-value of an exact permutation test. So you have strong evidence that the Likert ratings for the Multi-player game are higher.

In R, this can be formalized as a Fisher Exact Test on the $2\times 2$ data table TAB with rows for Single/Multi and columns for High/Low scores.

TAB = cbind(c(0,7),c(7,0)); TAB
     [,1] [,2]
[1,]    0    7
[2,]    7    0
fisher.test(TAB)

         Fisher's Exact Test for Count Data

data:  TAB
p-value = 0.0005828
alternative hypothesis: 
  true odds ratio is not equal to 1
95 percent confidence interval:
  0.0000000 0.1991716
sample estimates:
odds ratio 
         0 
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